AP Calculus AB Practice Quiz: Determining Concavity of Functions over Their Domains

Written by AP Content Team, Verified for 2026 AP Exams, Last updated: May 2026

Test your understanding with short quizzes. This quiz has 10 questions to check your progress.

Question 1 of 10

The graph of a function f(x) is concave up on an open interval I. Which of the following must be true for all x in I?

A)f'(x) > 0B)f'(x) < 0C)f''(x) > 0D)f''(x) < 0

All Questions (10)

The graph of a function f(x) is concave up on an open interval I. Which of the following must be true for all x in I?

A) f'(x) > 0

B) f'(x) < 0

C) f''(x) > 0

D) f''(x) < 0

Correct Answer: C

The second derivative of a function provides information about the function's concavity. A function is concave up on an interval where its second derivative is positive.

If the derivative of a function f(x), denoted by f'(x), is decreasing on the interval (a, b), which statement correctly describes the graph of f(x) on this interval?

A) The graph of f(x) is increasing.

B) The graph of f(x) is decreasing.

C) The graph of f(x) is concave up.

D) The graph of f(x) is concave down.

Correct Answer: D

According to the provided content, the graph of a function is concave down on an open interval if the function’s derivative is decreasing on that interval.

On which of the following intervals is the graph of the function f(x) = x³ - 6x² + 5 concave down?

A) (-∞, 2)

B) (2, ∞)

C) (-∞, 0)

D) (0, 4)

Correct Answer: A

To determine concavity, we find the second derivative. f'(x) = 3x² - 12x. f''(x) = 6x - 12. The graph is concave down when f''(x) < 0. Solving 6x - 12 < 0 gives 6x < 12, or x < 2. Therefore, the function is concave down on the interval (-∞, 2).

The graph of f'(x), the derivative of a function f(x), is shown. On which interval is the graph of f(x) concave up?

A) (-2, 1)

B) (1, ∞)

C) (-∞, -2)

D) (-∞, 1)

Correct Answer: B

The graph of f(x) is concave up on an interval where its derivative, f'(x), is increasing. Based on the provided graph of f'(x), the slope of the graph is positive (i.e., the function f'(x) is increasing) on the interval (1, ∞).

The function g(x) = x⁴ - 4x³ has points of inflection at which of the following x-values?

A) x = 0 only

B) x = 2 only

C) x = 0 and x = 2

D) x = 3 only

Correct Answer: C

A point of inflection occurs where the concavity changes, which corresponds to a sign change in the second derivative. First, find g''(x). g'(x) = 4x³ - 12x². g''(x) = 12x² - 24x. Set g''(x) = 0 to find potential inflection points: 12x(x - 2) = 0, which gives x = 0 and x = 2. We must test the sign of g''(x) around these points. For x < 0, g''(x) is positive. For 0 < x < 2, g''(x) is negative. For x > 2, g''(x) is positive. Since the sign of g''(x) changes at both x = 0 and x = 2, both are x-coordinates of points of inflection.

Let f be a function such that f''(x) exists for all x. If f''(c) = 0, which of the following statements must be true?

A) The graph of f has a point of inflection at x = c.

B) The graph of f has a local maximum or minimum at x = c.

C) The derivative of f, f'(x), has a local extremum at x = c.

D) The graph of f may have a point of inflection at x = c.

Correct Answer: D

The condition f''(c) = 0 identifies a candidate for a point of inflection. However, for a point of inflection to exist at x = c, the second derivative must change sign at x = c. For example, if f(x) = x⁴, then f''(x) = 12x², and f''(0) = 0, but f''(x) does not change sign at x=0, so there is no inflection point. Therefore, x=c is only a possible point of inflection.

For the function f(x) = xe⁻ˣ, the graph is concave down on which interval?

A) (-∞, 2)

B) (2, ∞)

C) (-∞, 1)

D) (1, ∞)

Correct Answer: A

We need to find where f''(x) < 0. Using the product rule, f'(x) = e⁻ˣ - xe⁻ˣ = e⁻ˣ(1-x). Using the product rule again, f''(x) = -e⁻ˣ(1-x) + e⁻ˣ(-1) = e⁻ˣ(-1+x-1) = e⁻ˣ(x-2). Since e⁻ˣ is always positive, the sign of f''(x) is determined by the sign of (x-2). The graph is concave down when f''(x) < 0, which occurs when x-2 < 0, or x < 2. So the interval is (-∞, 2).

The function f is twice-differentiable. Selected values of its derivative, f'(x), are given in the table below. | x | 0 | 1 | 2 | 3 | 4 | |---|---|---|---|---|---| | f'(x) | 5 | 3 | 2 | 3 | 5 | Based on the data, on which of the following intervals could the graph of f(x) be concave down?

A) (0, 2)

B) (2, 4)

C) (1, 3)

D) (3, 4)

Correct Answer: A

The graph of a function f(x) is concave down on an interval where its derivative, f'(x), is decreasing. Looking at the table, the values of f'(x) decrease from x=0 to x=1 (from 5 to 3) and from x=1 to x=2 (from 3 to 2). Therefore, f'(x) is decreasing over the interval (0, 2).

Suppose f is a function for which f''(x) > 0 for all x in the interval (a, b). Which of the following conclusions can be justified?

A) The function f is positive on (a, b).

B) The function f is increasing on (a, b).

C) The function f' (the derivative of f) is increasing on (a, b).

D) The function f' (the derivative of f) is positive on (a, b).

Correct Answer: C

The second derivative, f''(x), is the derivative of f'(x). If f''(x) > 0 on an interval, it means that the function of which it is the derivative, f'(x), is increasing on that interval. This, in turn, means that the original function f(x) is concave up. We cannot conclude anything about the sign of f(x) or f'(x).

Let h(x) = (1/12)x⁴ + (1/6)x³ - kx². For what value of k does the graph of h(x) have a point of inflection at x = -2?

A) k = -1

B) k = 1

C) k = 2

D) k = -2

Correct Answer: B

For a point of inflection to potentially exist at x = -2, we must have h''(-2) = 0. First, find the second derivative. h'(x) = (1/3)x³ + (1/2)x² - 2kx. h''(x) = x² + x - 2k. Now, set h''(-2) = 0: (-2)² + (-2) - 2k = 0, which simplifies to 4 - 2 - 2k = 0, or 2 - 2k = 0. Solving for k gives k = 1. We must also verify that the concavity changes at x = -2. With k=1, h''(x) = x²+x-2 = (x+2)(x-1). The sign of h''(x) changes from positive to negative at x=-2, confirming it is a point of inflection.